How do I find the point-normal form of the equation of the plane containing the point (-3,-4,3) and perpendicular to (4,1,-2)?

I know to use (r-p)*n)=0 but I keep getting 4x+y-2z=-10 and being told that's incorrect. Where am I going wrong? Have I just not simplified enough?

Thanks :)

1 Answer
Oct 2, 2017

The point-normal form of the equation of a plane is:

n_x(x-x_0)+ n_y(y-y_0)+ n_z(z-z_0) = 0

where < n_x, n_y, n_z > is the given normal vector and (x_0,y_0,z_0) is the given point.

Explanation:

Given the normal vector <4,1,-2> and the point (-3, -4, 3), the point-normal form is:

4(x-(-3))+ (y-(-4))-2(z-3) = 0

The above equation is in the requested point-normal form and it seems that you are trying to write the equation in the scalar form:

Ax+By+Cz = D

This is not the point-normal form.